Let $\langle A, R\rangle$ be a relational system in which $A$ has $n \neq 0$ elements. Let $R$ be connected, symmetric, and irreflexive on $A$. Thus, $R$ consists of all ordered pairs of distinct elements of $A$, i.e., all $\langle x, y\rangle$ for $x \neq y$. It is clear that $R$ contains $n(n-1)$ elements, since there are $n$ possibilities for $x$ and then $n-1$ possibilities for $y$ in $\langle x, y\rangle$. Obtain this result by a different method. Let $p(n)$ be the number of elements of $R$ when $A$ has $n$ elements. If $n=1$, then $p(n)=0$. Suppose we have $n$ elements in $A$ and $p(n)$ in $R$. Then we add a new element to $A$, so $A$ now has $n+1$ elements. How many new elements are added to $R$ ? Let $N$ be this number. Write a recursion formula for $p(n+1)$ in terms of $p(n)$ plus $N$, where $N$ is expressed in terms of $n$. Now use induction toE3.37. (Uses a common form of induction proof) prove that $p(n)=n(n-1)$. [Suggestion. After doing this exercise, go to the next one.]